This looks a lot like physics, but it is actually a math question! I will be omitting unnecessary constants for simplicity so the units might be off.
I want to reduce the equation
$-i\omega \vec{j}(\vec{q},\omega)=\vec{E}(\vec{q},\omega) +i\vec{q}\delta n(\vec{q},\omega)$
to an equation of the following form:
$\vec{j}(\vec{q},\omega)=\underline{\sigma}(\vec{q},\omega)\vec{E}(\vec{q},\omega)$
with current density $\vec{j}$, conductivity tensor $\underline{\sigma}$, $\vec{E}$ the electric field and wave vector $\vec{q}$. The factors $-i\omega$ and $i\vec{q}$ come frome the derivatives of the respective fourier transforms (once with respect to time and once with respect to $\vec{x}$).
It is given that $\delta n(\vec{q},\omega)=-\vec{q}\cdot \vec{j}\frac{1}{e\omega}$ where $e$ stands for the electron charge.
Now here comes the tricky part: In the solutions they plug in the given definition of $\delta n$ and then write the last term of the first equation as
$-i\frac{1}{e\omega}\vec{q}\vec{q}^T \cdot \vec{j}$ where $\vec{q}\vec{q}^T$ produces a matrix. My question is: where does this come from? Of course a matrix has to come out eventually (as I am explicitly looking for the conductivity tensor) but in my naivety I did the following:
$i\vec{q}\delta n(\vec{q},\omega)\rightarrow-i\frac{1}{e\omega }q^2 \cdot \vec{j}$
which I could also represent as a matrix multiplication
$-i\frac{1}{e\omega }q^2 \cdot \vec{j}=-i\frac{1}{e\omega }q^2 \mathbb{I} \vec{j}$
where $\mathbb{I}$ is supposed to stand for the identity matrix.
So, when multiplicating two vectors, when do I tensor and when do I not?